Constructing a circle tangent to another circle and two sides of a triangle

Question

Given the circle tangent to the sides $AB$ and $BC$, I want to construct another circle that is tangent to this circle and also tangent to the sides $AB$ and $AC$.

The center of such circle lies on the angle bisector of $\angle A$. Furthermore, the locus of the points that are equidistant from both $AB$ and the circle with center $O$ is a parabola with its focus at $O$ and the directrix parallel to $AB$ with the distance equal to the radius of the circle as mentioned here. Thus, the center of such circle can be found by intersecting the parabola and the angle bisector of $\angle A$ But I don't think it's possible to do this construction by using just a compass and a straight-edge.

Another thing that I've noticed is that when two circles are tangent to each other, their centers and their tangent point are placed on the same line. So, in order to do this construction it is enough to find the point of tangency and connect it to the center $O$ and extend it so that it intersects the angle bisector $A$ which gives the center of this circle. But I'm not sure how to find the point of tangency.

Answer 1

As noted in the comments and answers, this is a special case of the Problem of Apollonius, which dates back to antiquity. This special case is examined at cut-the-knot.org, from which this answer is adapted.

OP has already noted that the solution boils down to finding the point of tangency between the given and sought circles. What makes this possible is the observation that the point $S$ at which the circles touch would be a center of a homothety that takes either circle (and its tangents) to the other (and its tangents).

The construction is simple. We construct tangents to the given circle that are parallel to the sides $AC$ and $AB$. These tangents meet at $H$, and the line $AH$ consists of centers of homotheties that take tangents $AB$ and $AC$ of the sought circle to the constructed tangents of the given circle (and vice versa). There are two intersections of $AH$ with the given circle. One intersection is on $BC$ and leads to a trivial solution. But the other intersection $S$ gives us the point of contact between the two circles, and the intersection of $OS$ with the angle bisector of $\angle A$ gives us $K,$ the center of the sought circle.

Answer 2

Hint.

Considering the segment $s=[A,B]$ and calling $\alpha = \frac 12\angle \hat{BAC}$, the unknown radius $r$ and $x$ the distance over $s$ from $A$ to the vertical projection of the unknown circle's center, and $d_0$ the distance from $A$ over $s$ to the known circle's vertical projection, we have:

$$ (r+r_0)^2=(d_0-x)^2+(r-r_0)^2 $$

with $x = \frac{r}{\tan\alpha}$ and $r_0$ the known circle's radius, thus we can calculate $r$.

Answer 3

I am posting this as an answer because I was asked to by the inquirer

This is one of the Problem of Apollonius cases. Is the two lines and one circle case. Look here:

https://en.wikipedia.org/wiki/Problem_of_Apollonius

Answer 4

Your start and instinct were right, here is how to complete your construct (clockwise from upper left):

Draw the directrix $d$, bisect the angle of interest, from $O$ draw the perpendicular to the angle bisector with intersection $K$, mark off $KO'=KO$, and mark the intersection with the directrix as $J$. With diameter $JK$ create a circle centered on the line $JK$ (at the unmarked midpoint), and with diameter $OO'$ create a circle centered on line $OO'$ (at $K$), mark their intersection at $L$. With radius $JL$ and center $J$ draw a circle and mark the intersection with the directrix at $M$. Draw a perpendicular at $M$ and mark its intersection with the angle bisector at $N$.

By construction $JM=JL$, $JLK$ is a circle with diameter $JK$ so $\angle{JLK}$ is right and therefore $JL$ is tangent to circle $OLO'$ which has center $K$. Since $JL$ is a tangent and $JOO'$ a secant to circle $OLO'$, $JL^2=JO \times JO'$. But $JM=JL$ so $JM^2=JO \times JO'$. Now, by symmetry $NO=NO'$ so the circle with center $N$ and radius $NO$ has the secant $JOO'$, and since $JM^2=JO \times JO'$ it has tangent $JM$. Thus $NO=NM$, and $N$ is the intersection of the angle bisector and the parabola defined by focus $O$ and directrix $d$.

As others have commented, Apollonius tackled constructs like this in his Conics and some of his lost manuscripts. But the above demonstration depends only on Books 1-3 of Elements, and mainly on the proofs on tangents toward the end of Book 3.